Pascal's Triangle and the Binomial Theorem

Where we are headed#

$(a + b)^2$ and $(a + b)^3$ you already know by heart. But $(a + b)^{10}$? Nobody mul­ti­plies that out by hand. The good news is that its coef­fi­cients are sim­ply counts of ways to choose. We will build Pas­cal's tri­an­gle, prove the bino­mial the­o­rem for pos­i­tive inte­ger pow­ers, and then put it to work: expand­ing brack­ets, work­ing out pow­ers quickly, and decid­ing which of two large num­bers is big­ger.

Pas­cal's tri­an­gle#

$$\begin{array}{c} 1 \\ 1\quad 1 \\ 1\quad 2\quad 1 \\ 1\quad 3\quad 3\quad 1 \\ 1\quad 4\quad 6\quad 4\quad 1 \\ 1\quad 5\quad 10\quad 10\quad 5\quad 1 \end{array}$$

Look at any entry: it is the sum of the two num­bers just above it. That is Pas­cal's rule, $^nC_{r-1} + {}^nC_r = {}^{n+1}C_r$. Row $n$ sim­ply lists $^nC_0, {}^nC_1, \ldots, {}^nC_n$. Long before Pas­cal, Pin­gala knew this pat­tern and called it the Meru Pras­tara.

The bino­mial the­o­rem#

For a pos­i­tive inte­ger $n$,

$$\displaystyle (a + b)^n = \sum_{k=0}^{n} {}^nC_k\, a^{n-k} b^k = {}^nC_0 a^n + {}^nC_1 a^{n-1} b + \cdots + {}^nC_n b^n.$$

Why does this work? When you mul­ti­ply out the brack­ets, each term is made by pick­ing $a$ or $b$ from each of the $n$ brack­ets. So the term $a^{n-k}b^k$ turns up once for every way of choos­ing which $k$ brack­ets give you $b$, which means exactly $^nC_k$ times. (If you pre­fer a proof by induc­tion, Pas­cal's rule does the heavy lift­ing.)

Things worth notic­ing. You get $n + 1$ terms. The pow­ers of $a$ fall from $n$ to $0$ while the pow­ers of $b$ climb, and coef­fi­cients the same dis­tance from either end are equal. Put $a = b = 1$ and you get $\displaystyle \sum {}^nC_k = 2^n$. Take $a = 1$, $b = -1$ instead and you find $^nC_0 - {}^nC_1 + {}^nC_2 - \cdots = 0$.

Two cases you will use con­stantly. $\displaystyle (a - b)^n = \sum (-1)^k\, {}^nC_k\, a^{n-k}b^k$, and $\displaystyle (1 + x)^n = \sum {}^nC_k x^k$.

Putting it to work#

Try these your­self#

1. Write row $7$ of Pas­cal's tri­an­gle.
2. Expand: $(x + 2y)^5$; $(3 - x)^4$; $\displaystyle \left(\tfrac{x}{2} + \tfrac{2}{x}\right)^4$.
3. Use the the­o­rem to work out $(102)^5$; $(98)^4$; $(1.1)^5$.
4. Which is big­ger, $(1.01)^{100}$ or $2$?
5. Find $(x + 1)^6 + (x - 1)^6$ and use it to get $(\sqrt{2} + 1)^6 + (\sqrt{2} - 1)^6$.
6. Show that $9^{n+1} - 8n - 9$ is divis­i­ble by $64$ for every pos­i­tive inte­ger $n$.
7. Find the sum $^{10}C_0 + {}^{10}C_1 + \cdots + {}^{10}C_{10}$ and the sum of the even-posi­tion coef­fi­cients $^{10}C_0 + {}^{10}C_2 + \cdots + {}^{10}C_{10}$.
8. Expand $(1 + x + x^2)^3$ using $(1 + (x + x^2))^3$.

Answers to check against#